\(X=\left(a+b\right)^n=\sum\limits^n_{k=0}C^k_n.a^k.b^{n-k}\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

\(\Rightarrow A=\sum\limits^{90}_{k=2}C^k_{90}.2^k=...\)

Hoặc có thể làm như vầy: \(A=X-C^0_{90}.2^0-C^1_{90}.2=3^{90}-1-90.2=...\)

a) Ta có : \(\frac{-2}{x}=\frac{y}{3}\Leftrightarrow xy=-6\) 

Vì x < 0 < y nên

b) Ta có : \(\frac{x-3}{y-2}=\frac{3}{2}\)

\(\Leftrightarrow2\left(x-3\right)=3\left(y-2\right)\)

\(\Leftrightarrow2x-6=3y-6\)

\(\Leftrightarrow2x=3y\)

Mà x - y = 4 => x = 4 + y,do đó \(2\left(4+y\right)=3y\)

=> 8 + 2y = 3y

=> 3y - 2y = 8

=> y = 8

Thay y = 8 vào x - y = 4 ta có :

x - 8 = 4 => x = 4 + 8 = 12

Vậy x = 12,y = 8

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khó z tui chưa học mà :)

ta có 2+2^2=6>3

suy ra 2+2^2+2^3+...+2^90>C

a, (x+2)(x−2)−(x−3)(x+1)(x+2)(x-2)-(x-3)(x+1)

=x2−4−(x2−3x+x− 3)=x2-4-(x2-3x+x- 3)

=x2− 4−x2+2x+3=x2- 4-x2+2x+3

=2x−1=2x-1

2.

a, x2−4+(x−2)2x2-4+(x-2)2

=(x−2)(x+2) +(x−2)2=(x-2)(x+2) +(x-2)2

=(x−2)(x+2+x−2)=(x-2)(x+2+x-2)

=2x(x−2)=2x(x-2)

b, x3−2x2+x−xy2x3-2x2+x-xy2

=x(x2− 2x+1−y2)=x(x2- 2x+1-y2)

=x[(x−1)2 −y2]=x[(x-1)2 -y2]

=x(x−1−y)(x−1+y)=x(x-1-y)(x-1+y)

c, x3−4x2−12x+27x3-4x2-12x+27

=(x3+27)−(4x2+12x)=(x3+27)-(4x2+12x)

=(x+3)(x2−3x+9)−4x(x+3)=(x+3)(x2-3x+9)-4x(x+3)

=(x+3)(x2−3x+9−4x)=(x+3)(x2-3x+9-4x)

=(x+3)(x2−7x+9)